Question

Suppose that you work for Chick-fil-A and your manager has asked you to look at the wait times in the Drive-Thru. Supposing that wait times are normally distributed with a mean wait time of 30 seconds and a population standard deviation of 5 seconds. What is the probability that a randomly selected person will wait more than 42 seconds

Answer 1

0.0082 = 0.82% probability that a randomly selected person will wait more than 42 seconds.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean wait time of 30 seconds and a population standard deviation of 5 seconds.

This means that
`\mu = 30, \sigma = 5`

What is the probability that a randomly selected person will wait more than 42 seconds?

This is 1 subtracted by the pvalue of Z when X = 42. So

`Z = (X - \mu)/(\sigma)`

`Z = (42 - 30)/(5)`

`Z = 2.4`

`Z = 2.4` has a pvalue of 0.9918

1 - 0.9918 = 0.0082

0.0082 = 0.82% probability that a randomly selected person will wait more than 42 seconds.